Parcel Dropped From an Aircraft: A Step-by-Step Projectile Motion Solution (IB Physics)
A parcel released from a level-flying aircraft is one of the most common projectile motion questions. It asks for the path, the time to land, the vertical speed and the landing speed. Here is the whole thing solved step by step, including the rounding slip that quietly costs marks.
The problem
A small aircraft flies horizontally at 42 m/s at a height of 63 m above level ground. It releases a small parcel. Air resistance is negligible. Take g = 9.8 m/s². Find: (i) the path of the parcel, (ii) the time to reach the ground, (iii) the vertical component of its velocity just before landing, (iv) its speed on landing.
This is an exam-style problem described in our own words with our own numbers worked from it, not a copy of any paper.
Part (i): sketch the path
The path starts horizontal, because at release the parcel is moving only at 42 m/s along the flight direction. It then curves downward, getting steeper, and lands ahead of the release point. It is a curve (a parabola), not a straight line down and not a straight slope.
Part (ii): time to reach the ground
Use the vertical axis only. The initial vertical velocity is 0 because the aircraft is level, and the acceleration is g = 9.8 m/s² downward.
s = ut + ½at² gives 63 = ½ × 9.8 × t². So 63 = 4.9t², t² = 12.86 and t = 3.59 s (about 3.6 s). The aircraft speed of 42 m/s was not needed. Horizontal speed never changes how long it takes to fall.
Part (iii): vertical speed just before landing
v = u + at on the vertical axis: v_y = 0 + 9.8 × 3.586 = 35.1 m/s downward. A quick check that does not need the time: v_y = √(2gh) = √(2 × 9.8 × 63) = 35.1 m/s.
Part (iv): speed on landing
The horizontal component is still 42 m/s, because nothing acted sideways. The two components are at right angles, so use Pythagoras: speed = √(42² + 35.1²) = 54.7 m/s, at about 40° below the horizontal.
You can check it with energy. The gravitational potential energy lost equals the kinetic energy gained, so v² = 42² + 2 × 9.8 × 63 = 2998.8, which gives 54.8 m/s. The mass cancels, so a heavier parcel lands at the same speed.
The rounding trap
If you round the time to 3.6 s and then use it again, you get v_y = 9.8 × 3.6 = 35.3 m/s instead of 35.1 m/s. The 54.7 and 54.8 m/s answers differ for a similar reason: 35.1 was rounded before squaring. The habit that protects your marks is to keep the unrounded value in the calculator and round only the final answer.
Where is the aircraft when the parcel lands?
If the aircraft carries on at a steady 42 m/s and air resistance is ignored, it is directly above the parcel all the way down, because both have the same horizontal velocity. The parcel lands about 42 × 3.586 = 151 m ahead of where it was released.
Frequently asked questions
No. The time of fall depends only on the height and g. A faster aircraft makes the parcel land further away, not later.
Find the horizontal and vertical components just before landing, then combine them with Pythagoras: speed = √(v_x² + v_y²). You can check it with energy: v² = u² + 2gh for a horizontal launch.
Rounding. If you round the vertical speed to 35.1 m/s before squaring you get 54.7 m/s; with unrounded values you get 54.8 m/s. Both round to 55 m/s, so keep extra digits in the working.
Directly above it, because the aircraft and the parcel keep the same horizontal velocity when air resistance is ignored.